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A little bit earlier I posted this question about the equation using the $\gamma$ Euler-Mascheroni constant. There I kinda understood what was going on with help of fellow MSE colleagues, so I decided to test it on the following limit.

$$\lim_{n\to\infty}\left(\frac{1}{2n+1}+\cdots+\frac{1}{9n}\right)$$

The mentioned equation is as follows:

$1+\frac{1}{2}+\cdots+\frac{1}{n}=\gamma+\mathcal{E}_n+\ln(n)\space,\ \mathcal{E}_n\longrightarrow 0\space\ when\space\ n\longrightarrow\infty$

By the same procedure as in the mentioned question,

$$1+\frac{1}{2}+\cdots+\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}+\frac{1}{2n+1}+\cdots+\frac{1}{9n}-\left(1+\frac{1}{2}+\cdots+\frac{1}{2n}\right)= \\ =/H_n=\gamma+\mathcal{E}_n+\ln(n)/= \\ =\gamma+\mathcal{E}_{9n}+\ln(9n)-(\gamma+\mathcal{E}_{2n}+\ln(2n))=\mathcal{E}_{9n}-\mathcal{E}_{2n}+\ln\left(\frac{9}{2}\right)$$

$$\text{And thereby,}$$

$$\lim_{n \to \infty}\left(\frac{1}{2n+1}+\cdots+\frac{1}{9n}\right)=\lim_{n \to \infty}\underbrace{\mathcal{E}_{9n}}_{\rightarrow 0}-\underbrace{\mathcal{E}_{2n}}_{\rightarrow 0}+\ln\left(\frac{9}{2}\right)=\ln\left(\frac{9}{2}\right)$$

My question, is everything alright with this ?

I find the method very nice and furthermore interesting since I haven't been seeing it a lot around here.

Thanks

powerline
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    GrigoriPerelman, your proof is correct. Well done :) – LHF Jan 27 '20 at 07:11
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    It is alright. But you could use some line breaks instead of putting the main parts in one line :-) – trancelocation Jan 27 '20 at 07:11
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    You can get the proper font for $\ln$ by using \ln. For operators without a command of their own, you can use \operatorname{name}. To prevent text such as "when" within equations from being italicized, you can use \text{when}. – joriki Jan 27 '20 at 07:25
  • @trancelocation I broke some lines. I am on phone, when I get to pc I'll check it again – powerline Jan 27 '20 at 07:28
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    @joriki all lns repaired. Actually, in some cases I keep the words italicized on purpose like this one. Since it fits better into the look of the whole line, it's almost as it's an operator constituted from letters and not a word. Thanks anyway for the useful information – powerline Jan 27 '20 at 07:36
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    In fact $\mathcal{E}_n\sim\frac{1}{2n}$, but you don't need that. – J.G. Jan 27 '20 at 10:54
  • @J.G. I am really interested in that, tell me more please. Thanks – powerline Jan 27 '20 at 15:02
  • @GrigoriPerelman Sure. – J.G. Jan 27 '20 at 15:06
  • @J.G. I suppose the tilde represents a symmetry relation between those two. Though, I've been coming across that relation only when describing some basic equivalence relations, and therefore I am not sure I understand what it means here. Could you elaborate, just in few words. Thanks for the link, I see you commented there just 2 days ago :-) – powerline Jan 27 '20 at 15:15
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    @GrigoriPerelman It means $\lim_{n\to\infty}2n\mathcal{E}_n=1$. – J.G. Jan 27 '20 at 15:19

3 Answers3

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Yes, you are right.

Another way:

$$\lim_{n\rightarrow+\infty}\sum_{k=2n+1}^{9n}\frac{1}{k}=\lim_{n\rightarrow+\infty}\frac{1}{n}\sum_{k=1}^{7n}\frac{1}{2+\frac{k}{n}}=\int_0^7\frac{1}{2+x}dx=\ln4.5.$$

Michael Rozenberg
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You are correct.

Next, as an exercise, prove that, using the $H_n$ approximation, if $a, b, c, d$ are integers with $0 < a < c$ then $\lim_{n \to \infty} \sum_{k=an+b}^{cn+d} \dfrac1{k} =\ln(\frac{c}{a}) $.

Then try with $a, b, c, d$ reals instead of integers.

marty cohen
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Asm marty cihen suggested, harmonic numbers are interesting for the limit and more than.

You problem is $$S_n=\sum_{i=1}^{7n}\frac 1{2n+i}=H_{9 n}-H_{2 n}$$

For large $n$, remember that $$H_p=\left(\gamma +\log \left({p}\right)\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^3}\right)$$ If you just use the term in brackets, the limit is immediate.

If you want to go further, apply twice the formula to get $$S_n=(\log (9)-\log (2))-\frac{7}{36 n}+\frac{77}{3888 n^2}+O\left(\frac{1}{n^3}\right)$$

Try it for $n=5$. The exact value is $$S_5=\frac{13808926545210682009}{9419588158802421600}\approx 1.4659799 $$ while the trucated series gives $$S_5 \sim \log \left(\frac{9}{2}\right)-\frac{3703}{97200}\approx 1.4659807$$

Claude Leibovici
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